In: Statistics and Probability
6. A standard drug is used to treat a certain disease. The probability with which the drug is effective is 0.85. A new drug is developed and it is desired to determine if the new drug performs better than the standard. An experiment was conducted with which 300 people are given the new drug.
a. State symbolically the null and alternative hypotheses.
b. What is the critical value and rejection region of the proposed test?
c. If in fact, the success rate of the drug is 0.9, what is the power of the test?
d. When the experiment was actually carried out, it was found that the new drug was effective in 269 of the patients. Find the p-value of the test.
a)
p_1 be the effective rate of drug 1 = 0.85
p_2 be the effective rate of drug 2
null hypotheses : p_1 >= p_2
alternative hypotheses : p_1 < p_2 ( if new drug performs better )
b) Rejection region will be if p value for the test is greater than alpha ( 0.05 ), if critical region is beyond 0.05
c)
sample proportions , p = 0.9
SE(p) = sqrt( p*(1-p)/n ) = sqrt( 0.85*0.15/ 300 ) = sqrt(0.000425 ) = 0.0206
we need to find p_critical
P[ Reject H0 when it is true ] = alpha = 0.05
P[ Z > z_critical ] = 0.975 ( two tailed )
z_critical = 1.96
P[ ( p - P )/SE(p) > ( p_critical - P )/SE(p) ] = P[ ( p - 0.85 )/0.0206 > ( p_critical - 0.85 )/0.0206 ] =0.975
( p_critical - 0.85 )/0.0206 = 1.96
( p_critical - 0.85 ) = 0.04037
p_critical = 0.85 + 0.04037 = 0.89037
we will reject H0 if p is more than 0.89037
P[ Accepting H0 when it is true ] = P[ p < p_critical ] = P[ ( p - 0.9 )/0.0206 < ( 0.89037 - 0.9 )/0.0206 ] = P[ Z < -0.4674 ] = 0.3203
d) sample size , n = 300
new drug was effective for 269 patients
sample proportions , p = 269/300 = 0.8967
SE(p) = sqrt( p*(1-p)/n ) = sqrt( 0.85*0.15/ 300 ) = sqrt(0.000425 ) = 0.0206
z = ( p - P )/SE(p) = ( 0.8967 - 0.85 )/0.0206 = 2.2667
p value = 0.9879
we reject H0